A projectile fired with a speed of v=60m/s at an angle of 60°. a second projectile is then fired with the same initial speed 0.5s later. Find the angle theta of the second projectile so that they will collide and what position (x, y) will it occur?

1 Answer
Jun 12, 2018

The angle is theta=57.58^@ and the position is (222.6, 116.1)

Explanation:

The horizontal distance travelled is

{(x=vcos60t),(x=vcostheta(t-0.5)):}

Therefore,

60*1/2*t=60costheta(t-0.5)

30t=60(t-0.5)costheta

t=2tcostheta-costheta

t(2costheta-1)=costheta

t=costheta/(2costheta-1)

The vertical distance travelled is

{(y=vsin60*t-1/2g t^2=52t-4.9t^2),(y=vsintheta(t-0.5)-1/2g(t-0.5)^2=60tsintheta-30sintheta-4.9(t-0.5)^2):}

{(y=52t-4.9t^2),(y=60tsintheta-30sintheta-4.9t^2+4.9t-1.23):}

Therefore,

52t-4.9t^2=60tsintheta-30sintheta-4.9t^2+4.9t-1.23

60tsintheta-30sintheta+4.9t-52t-1.23=0

t(60sintheta-47.1)=30sintheta+1.23

t=(30sintheta+1.23)/(60sintheta-47.1)

The t's must be equal

(30sintheta+1.23)/(60sintheta-47.1)=costheta/(2costheta-1)

(30sintheta+1.23)(2costheta-1)=costheta(60sintheta-47.1)

60sinthetacostheta-30sintheta+2.46costheta-1.23=60sintheta costheta-47.1costheta

-30sintheta+49.56costheta=1.23

49.56costheta-30sintheta=1.23

Solving for theta

rsin(alpha-theta)=rsinalphacostheta-rsinthetacosalpha

{(rcosalpha=-30),(rsinalpha=49.56):}

<=>, r^2=sqrt(30^2+49.56^2)=57.93

{(cosalpha=-30/57.93=-0.558),(sinalpha=49.56/57.93=0.83):}

alpha=58.8^@

Therefore,

57.93sin(58.8-theta)=1.23

sin(58.8-theta)=1.23/57.93=0.021

58.8-theta=1.22

theta=58.8-1.22=57.58^@

And

t=cos(57.6)/(2cos(57.6)-1)=7.42s

x=60cos(57.58)(7.42-0.5)=222.6m

y=52*7.42-4.9*7.42^2=116.1m